Perm / Combi

[govind_raj_76]

How many 4 digit numbers that are divisible by 2 can be formed so that only the two rightmost digits are even and the digit 0 cannnot appear more than once ?

A) 300

B) 400

C) 500

D) 600

E) 700

Please explain the best approach to solve this problem ?

Thanks,

[clock60]

How many 4 digit numbers that are divisible by 2 can be formed so that only the two rightmost digits are even and the digit 0 cannnot appear more than once ?

A) 300

B) 400

C) 500

D) 600

E) 700

Please explain the best approach to solve this problem ?

Thanks,

i got D-600, reasoning
if i got the problem right the desired number must look like
odd odd even even and 0 used only once
for me all numbers will end 0,2,4,6,8

odd odd even 0=5*5*4*1=100 (4 -as we chose from 2,4,6,8 and 0 is already chosen)
odd odd even 2=5*5*5*1=125 ( here we have 0,2,4,6,8)
odd odd even 4=5^3=125
odd odd even 6=125
odd odd even 8=125

total=100+125*4=600

[iamseer]

IMO answer D

_ _ _ _

5*5*(4*1 +1*4+4*4) =600

Now, since only last 2 digits are even, the first 2 digits are odd.
First digit can be selected in 5 ways (1,3,5,7,9)
Second digit can be selected in 5 ways ((1,3,5,7,9)

Since 0 is not to be repeated, it can be at ten's place, unit's place or nowhere

0 in unit's place - Ten's place can be selected in 4 ways (2,4,6,8)
0 in ten's place - Unit's place can be selected in 4 ways (2,4,6,8)
0 nowhere - Unit's place and Ten's place can both be selected in 4 ways each

HTH

[susantaiitk]

first 2 places are odd [ 1,3,5,7,9] total ways = 5*5 = 25
last 2 places are even [ 0,2,4,6,8] total ways = 5*5 = 25

but 0 can't be more than once, hence last 2 digit 00 is avoided then total ways of last 2 places = 25 - 1= 24

total ways = 25*24 = 600

[sumanr84]

first 2 places are odd [ 1,3,5,7,9] total ways = 5*5 = 25
last 2 places are even [ 0,2,4,6,8] total ways = 5*5 = 25

but 0 can't be more than once, hence last 2 digit 00 is avoided then total ways of last 2 places = 25 - 1= 24

total ways = 25*24 = 600

Fastest...Simple..clear and best solution..

[govind_raj_76]

How many 4 digit numbers that are divisible by 2 can be formed so that only the two rightmost digits are even and the digit 0 cannnot appear more than once ?

A) 300

B) 400

C) 500

D) 600

E) 700

Please explain the best approach to solve this problem ?

Thanks,

i got D-600, reasoning
if i got the problem right the desired number must look like
odd odd even even and 0 used only once
for me all numbers will end 0,2,4,6,8

odd odd even 0=5*5*4*1=100 (4 -as we chose from 2,4,6,8 and 0 is already chosen)
odd odd even 2=5*5*5*1=125 ( here we have 0,2,4,6,8)
odd odd even 4=5^3=125
odd odd even 6=125
odd odd even 8=125

total=100+125*4=600

Nice Explaination !!

[debmalya_dutta]

Is the question missing the portion which applies the constraint that the 2 left most digits should be odd digits ? Strictly based on the question , the 1000 and the 100's place can have both even and odd digits.. why have you guys considered only 5 ways